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Student Study Guide and Solutions Manual, 4e for Organic Chemistry, 4e David Klein Johns Hopkins University SVP, PUBLISHING STRATEGY AND QUALITY: Elizabeth Widdicombe ASSOCIATE PUBLISHER: Sladjana Bruno MARKETING MANAGER: Michael Olsen SENIOR MANAGING EDITOR: Mary Donovan EXECUTIVE MANAGING EDITOR: Valerie Zaborski EDITORIAL ASSISTANT: Samantha Hart COURSE CONTENT DEVELOPER: Andrew Moore SENIOR COURSE PROUCTION OPERATIONS SPECIALIST: Patricia Gutierrez SENIOR MANAGER, COURSE DEVELOPMENT AND PRODUCTION: Svetlana Barskaya ART DIRECTOR/COVER DESIGNER: Wendy Lai Cover: Wiley Cover Images: Abstract geometric texture ©bgblue/Getty Images, Flask – Norm Christiansen, Colorful Paintbrushes © Maartje van Caspel/Getty Images, Honeycomb slice © eli_asenova/Getty Images, Color diffuse © Korolkoff/Getty Images, Rosemary © Tetiana Rostopira/Getty Images Founded in 1807, John Wiley & Sons, Inc has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support For more information, please visit our website: www.wiley.com/go/citizenship Copyright © 2021, 2017, 2015, 2012 John Wiley & Sons, Inc All rights reserved No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per‐copy fee to the Copyright Clearance Center, Inc 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030‐5774, (201)748‐6011, fax (201)748‐6008, website http://www.wiley.com/go/permissions Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year These copies are licensed and may not be sold or transferred to a third party Upon completion of the review period, please return the evaluation copy to Wiley Return instructions and a free of charge return shipping label are available at www.wiley.com/go/return label Outside of the United States, please contact your local representative ISBN: 978‐1‐119‐65958‐7 Printed in the United States of America 10 The inside back cover will contain printing identification and country of origin if omitted from this page In addition, if the ISBN on the back cover differs from the ISBN on this page, the one on the back cover is correct CONTENTS Chapter – Electrons, Bonds, and Molecular Properties Chapter – Molecular Representations 27 Chapter – Acids and Bases 69 Chapter – Alkanes and Cycloalkanes 110 Chapter – Stereoisomerism 141 Chapter – Chemical Reactivity and Mechanisms 173 Chapter – Alkyl Halides: Nucleophilic Substitution and Elimination Reactions 195 Chapter – Addition Reactions of Alkenes 262 Chapter – Alkynes 317 Chapter 10 – Radical Reactions 368 Chapter 11 – Synthesis 409 Chapter 12 – Alcohols and Phenols 449 Chapter 13 – Ethers and Epoxides; Thiols and Sulfides 510 Chapter 14 – Infrared Spectroscopy and Mass Spectrometry 563 Chapter 15 – Nuclear Magnetic Resonance Spectroscopy 593 Chapter 16 – Conjugated Pi Systems and Pericyclic Reactions 639 Chapter 17 – Aromatic Compounds 682 Chapter 18 – Aromatic Substitution Reactions 717 Chapter 19 – Aldehydes and Ketones 790 Chapter 20 – Carboxylic Acids and Their Derivatives 865 Chapter 21 – Alpha Carbon Chemistry: Enols and Enolates 929 Chapter 22 – Amines 1010 Chapter 23 – Introduction to Organometallic Compounds 1071 Chapter 24 – Carbohydrates 1126 Chapter 25 – Amino Acids, Peptides, and Proteins 1154 Chapter 26 – Lipids 1183 Chapter 27 – Synthetic Polymers 1202 HOW TO USE THIS BOOK Organic chemistry is much like bicycle riding You cannot learn how to ride a bike by watching other people ride bikes Some people might fool themselves into believing that it’s possible to become an expert bike rider without ever getting on a bike But you know that to be incorrect (and very naïve) In order to learn how to ride a bike, you must be willing to get on the bike, and you must be willing to fall With time (and dedication), you can quickly train yourself to avoid falling, and to ride the bike with ease and confidence The same is true of organic chemistry In order to become proficient at solving problems, you must “ride the bike” You must try to solve the problems yourself (without the solutions manual open in front of you) Once you have solved the problems, this book will allow you to check your solutions If, however, you don’t attempt to solve each problem on your own, and instead, you read the problem statement and then immediately read the solution, you are only hurting yourself You are not learning how to avoid falling Many students make this mistake every year They use the solutions manual as a crutch, and then they never really attempt to solve the problems on their own It really is like believing that you can become an expert bike rider by watching hundreds of people riding bikes The world doesn’t work that way! The textbook has thousands of problems to solve Each of these problems should be viewed as an opportunity to develop your problem-solving skills By reading a problem statement and then reading the solution immediately (without trying to solve the problem yourself), you are robbing yourself of the opportunity provided by the problem If you repeat that poor study habit too many times, you will not learn how to solve problems on your own, and you will not get the grade that you want Why so many students adopt this bad habit (of using the solutions manual too liberally)? The answer is simple Students often wait until a day or two before the exam, and then they spend all night cramming Sound familiar? Unfortunately, organic chemistry is the type of course where cramming is insufficient, because you need time in order to ride the bike yourself You need time to think about each problem until you have developed a solution on your own For some problems, it might take days before you think of a solution This process is critical for learning this subject Make sure to allot time every day for studying organic chemistry, and use this book to check your solutions This book has also been designed to serve as a study guide, as described below WHAT’S IN THIS BOOK This book contains more than just solutions to all of the problems in the textbook Each chapter of this book also contains a series of exercises that will help you review the concepts, skills and reactions presented in the corresponding chapter of the textbook These exercises are designed to serve as study tools that can help you identify your weak areas Each chapter of this solutions manual/study guide has the following parts: • • • • • • • Review of Concepts These exercises are designed to help you identify which concepts are the least familiar to you Each section contains sentences with missing words (blanks) Your job is to fill in the blanks, demonstrating mastery of the concepts To verify that your answers are correct, you can open your textbook to the end of the corresponding chapter, where you will find a section entitled Review of Concepts and Vocabulary In that section, you will find each of the sentences, verbatim Review of Skills These exercises are designed to help you identify which skills are the least familiar to you Each section contains exercises in which you must demonstrate mastery of the skills developed in the SkillBuilders of the corresponding textbook chapter To verify that your answers are correct, you can open your textbook to the end of the corresponding chapter, where you will find a section entitled SkillBuilder Review In that section, you will find the answers to each of these exercises Review of Reactions These exercises are designed to help you identify which reagents are not at your fingertips Each section contains exercises in which you must demonstrate familiarity with the reactions covered in the textbook Your job is to fill in the reagents necessary to achieve each reaction To verify that your answers are correct, you can open your textbook to the end of the corresponding chapter, where you will find a section entitled Review of Reactions In that section, you will find the answers to each of these exercises Review of Mechanisms These exercises are designed to help you practice drawing the mechanisms To verify that you have drawn the mechanism correctly, you can open your textbook to the corresponding chapter, where you will find the mechanisms appearing in numbered boxes throughout the chapter In those numbered boxes, you will find the answers to each of these exercises Common Mistakes to Avoid This is a new feature to this edition The most common student mistakes are described, so that you can avoid them when solving problems A List of Useful Reagents This is a new feature to this edition This list provides a review of the reagents that appear in each chapter, as well as a description of how each reagent is used Solutions At the end of each chapter, you’ll find detailed solutions to all problems in the textbook, including all SkillBuilders, conceptual checkpoints, additional problems, integrated problems, and challenge problems The sections described above have been designed to serve as useful tools as you study and learn organic chemistry Good luck! David Klein Johns Hopkins University Chapter A Review of General Chemistry: Electrons, Bonds and Molecular Properties Review of Concepts Fill in the blanks below To verify that your answers are correct, look in your textbook at the end of Chapter Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulary _ isomers share the same molecular formula but have different connectivity of atoms and different physical properties Second-row elements generally obey the _ rule, bonding to achieve noble gas electron configuration A pair of unshared electrons is called a A formal charge occurs when atoms not exhibit the appropriate number of _ An atomic orbital is a region of space associated with , while a molecular orbital is associated with _ Methane’s tetrahedral geometry can be explained using four degenerate _-hybridized orbitals to achieve its four single bonds Ethylene’s planar geometry can be explained using three degenerate _-hybridized orbitals Acetylene’s linear geometry is achieved via _-hybridized carbon atoms The geometry of small compounds can be predicted using valence shell electron pair repulsion (VSEPR) theory, which focuses on the number of bonds and _ exhibited by each atom The physical properties of compounds are determined by forces, the attractive forces between molecules London dispersion forces result from the interaction between transient and are stronger for larger alkanes due to their larger surface area and ability to accommodate more interactions Review of Skills Fill in the blanks and empty boxes below To verify that your answers are correct, look in your textbook at the end of Chapter The answers appear in the section entitled SkillBuilder Review SkillBuilder 1.1 Drawing Constitutional Isomers of Small Molecules SkillBuilder 1.2 Drawing the Lewis Structure of a Small Molecule CHAPTER SkillBuilder 1.3 Calculating Formal Charge SkillBuilder 1.4 Locating Partial Charges Resulting from Induction SkillBuilder 1.5 Reading Bond-Line Structures SkillBuilder 1.6 Identifying Electron Configurations Step In the energy diagram shown here, draw the electron configuration of nitrogen (using arrows to represent electrons) 2p 2s 1s Step Fill in the boxes below with the numbers that correctly describe the electron configuration of nitrogen 1s Nitrogen SkillBuilder 1.7 Identifying Hybridization States 2s 2p 54 CHAPTER nitrogen atom or 2) generate a nitrogen atom with less than an octet of electrons, or 3) generate a structure with three charges The first of these would not be a valid resonance structure, and the latter two would not be significant resonance structures (e) The condensed structure indicates how the atoms are connected to each other In the bond-line structure, hydrogen atoms are not drawn (they are implied) Each corner and each endpoint represents a carbon atom, so the carbon skeleton is shown more clearly 2.57 The negatively charged oxygen atom has three lone pairs, the positively charged oxygen atom has one lone pair, and the uncharged oxygen atom has two lone pairs (see Table 2.2) Notice that this compound exhibits a lone pair that is next to a bond, so we must draw two curved arrows associated with that pattern The first curved arrow is drawn showing a lone pair becoming a bond, while the second curved arrow shows a bond becoming a lone pair: (f) The condensed structure indicates how the atoms are connected to each other In the bond-line structure, hydrogen atoms are not drawn (they are implied) Each corner and each endpoint represents a carbon atom, so the carbon skeleton is shown more clearly These two resonance forms have completely filled octets and are equivalent (they contribute equally to the resonance hybrid) There are no other valid resonance structures that are significant 2.58 Each nitrogen atom has a lone pair that is adjacent to a bond and is, therefore, delocalized via resonance In order to be delocalized via resonance, the lone pair must occupy a p orbital, and therefore, each nitrogen atom must be sp2 hybridized As such, each nitrogen atom is trigonal planar 2.56 The nitronium ion does not have any significant resonance structures because any attempts to draw a resonance structure will either 1) exceed an octet for the 2.59 (a) This compound (estradiol) exhibits a lone pair next to a bond, so we draw two curved arrows The first curved arrow is drawn showing a lone pair becoming a bond, while the second curved arrow shows a bond becoming a lone pair We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns In this case, the lone pair is next to another bond, so once again, we draw the two curved arrows associated with that pattern The resulting resonance structure again exhibits a lone pair next to a bond, so again we draw two curved arrows and the resulting resonance structure Once again, there is a lone pair next to a bond, which requires that we draw one final resonance structure, shown below This last resonance structure is not the same as the original resonance structure, because of the locations in which the bonds are drawn (i.e., the resonance pattern of conjugated bonds enclosed in a ring) Notice that all resonance structures have zero net charge: OH OH H H H OH H H H H HO HO H H HO OH OH H H HO H H H HO H CHAPTER 55 (b) The following compound (testosterone) exhibits a C=O bond (a bond between two atoms of differing electronegativity), so we draw one curved arrow showing the bond becoming a lone pair on the more electronegative oxygen atom We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns In this case, there is a C+ (carbocation) that is allylic, so we draw the curved arrow associated with that pattern (pushing over the bond) Notice that all resonance structures have zero net charge: 2.60 This compound exhibits a C=O bond (a bond between two atoms of differing electronegativity), so we draw one curved arrow showing the bond becoming a lone pair on the more electronegative oxygen atom We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns In this case, there is an allylic C+ (carbocation), so we draw the curved arrow associated with that pattern (pushing over the bond), shown here This pattern continues, several more times, spreading the positive charge over a total of six carbon atoms Notice that all resonance structures have zero net charge: By considering the significant resonance structures (drawn above), we can determine the positions that are electron deficient (+) There are six sites of low electron density: 2.61 This compound exhibits a lone pair next to a bond, so we draw two curved arrows The first curved arrow is drawn showing a lone pair becoming a bond, while the second curved arrow shows a bond becoming a lone pair We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns In this case, there is a lone pair next to a bond, so once again, we draw the two curved arrows associated with that pattern The resulting resonance structure again exhibits a lone pair next to a bond This pattern continues, several more times, spreading a negative charge over a total of five carbon atoms Notice that all resonance structures have zero net charge: 56 CHAPTER By considering the significant resonance structures (drawn above), we can determine the positions that are electron rich (‒) There are five sites of high electron density: 2.62 Two patterns of resonance can be identified on the given structure: carbonyl resonance and allylic lone pair resonance (involving either the oxygen or nitrogen lone pairs) All three of these options will be used Since it is possible to start with any one of the three, you may have developed the resonance forms in a different order than presented here, but you still should have found four reasonable resonance forms Only one of the four structures has an atom with an incomplete octet (the last resonance structure shown), so that is identified as the least significant contributor to the hybrid The first resonance form is the most significant contributor because it has filled octets and no formal charges The middle two resonance structures both have filled octets and a negative charge on an oxygen atom, so they are ranked according to their only difference: the location of the positive charge The third structure is the 2nd most significant resonance form because it has the positive charge on the less electronegative nitrogen atom Note it is better to place a negative charge on a more electronegative atom, and it is better to place a positive charge on a less electronegative atom 2.63 The only resonance pattern evident in the enamine is an allylic lone pair After that pattern is applied, however, another allylic lone pair results so the resonance can ultimately involve both bonds There are a total of three major resonance forms that all have filled octets Consideration of the hybrid of these resonance forms predicts two electronrich sites (‒) Did you draw the following additional structure (or something similar, with C+ and C) and wonder why it was not shown in this solution? This resonance form suffers from two major deficiencies: 1) it does not have filled octets, while the other resonance forms shown above all have filled octets, and 2) it has a negative charge on a carbon atom (which is not an electronegative atom) Either of these deficiencies alone would render the resonance form a minor contributor But with both deficiencies together (C+ and C–), this resonance form is insignificant The same is true for any resonance form that has both C+ CHAPTER 57 and C– Such a resonance form will generally be insignificant (there are very few exceptions, one of which will be seen in Chapter 17) Also, note that the bonds cannot be moved to other parts of the six-membered ring since the CH2 groups are sp3 hybridized These carbon atoms cannot accommodate an additional bond without violating the octet rule 2.64 The correct answer is (c) Structures (a), (b) and (d) are all significant resonance structures, as shown: Specifically, this resonance structure has alternating single and double bonds in the ring and is considered to be aromatic (a stabilizing effect) 2.66 The correct answer is (d) The nitrogen atom in structure (a) is delocalized by resonance, and is therefore sp2 hybridized: Structure (c) is not a resonance form at all To see this more clearly, notice that the benzyl carbocation does not have any CH2 groups in the ring, but structure (c) does have a CH2 group in the ring: Resonance structures differ only in the placement of electrons Since structure (c) differs in the connectivity of atoms, it cannot be considered a resonance structure of the benzyl carbocation 2.65 The correct answer is (a) The atoms in all four structures have complete octets So we must consider the location of the negative charge Structure (a) has a negative charge on an electronegative atom (oxygen) A negative charge is more stable on the more electronegative atom (oxygen) than it is on a nitrogen atom or a carbon atom Therefore, structure (a) will contribute the most character to the resonance hybrid: The nitrogen atom in structure (b) is also delocalized, so it is sp2 hybridized as well: The nitrogen atom in structure (c) is also sp2 hybridized, because this nitrogen atom must be using a p orbital to participate in bonding (C=N) The nitrogen atom in structure (d) has three bonds, and its lone pair is localized Therefore, this nitrogen atom is sp3 hybridized: 2.67 The correct answer is (c) We begin by considering all bonds in the compound, which is easier to if we redraw the compound as shown: This compound corresponds structure (c): It turns out that this resonance structure is also significant for another reason that we will explore in Chapter 17 58 CHAPTER 2.68 The correct answer is (c) For C+ that is allylic, only one curved arrow will be required to draw the resonance structures (pushing over the bond): 2.69 The correct answer is (a) An anion will be resonance delocalized if the atom bearing the negative charge has a lone pair that is next to a π bond Only compound I fits this pattern: 2.70 The correct answer is (a) There are only two resonance structures possible for this allylic carbocation Drawing all the hydrogen atoms helps to demonstrate why the cation cannot be delocalized to any other position without breaking a single bond (and we cannot break a single bond when drawing resonance structures) 2.71 The correct answer is (d) Begin by drawing all significant resonance structures In this case, there are two: Both are equally significant, so the resonance hybrid is the simple average of these two resonance structures When drawing the resonance hybrid, partial bonds and partial charges are required, as shown below: 2.72 The correct answer is (b) The following not represent a pair of resonance structures, because the first rule of resonance has been violated (by breaking a single bond): 2.73 (a) The molecular formula is C3H6N2O2 (b) Each of the highlighted carbon atoms (below) has four sigma bonds (the bonds to hydrogen are not shown) As such, these two carbon atoms are sp3 hybridized (c) There is one carbon atom that is using a p orbital to form a bond As such, this carbon atom (highlighted) is sp2 hybridized (d) There are no sp hybridized carbon atoms in this structure (e) There are six lone pairs (each nitrogen atom has one lone pair and each oxygen atom has two lone pairs): (f) Only the lone pair on one of the nitrogen atoms is delocalized via resonance (because it is next to a bond) The other lone pairs are all localized (g) The geometry of each atom is shown below (see SkillBuilder 1.8): CHAPTER (h) We begin by looking for the five patterns, focusing first on any patterns that employ just one curved arrow (in this case, there is another pattern that requires two curved arrows, but we will start with the pattern using just one curved arrow) There is a C=O bond (a bond between two atoms of differing electronegativity), so we draw one curved arrow showing the bond becoming a lone pair on the more electronegative oxygen atom We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns In this case, there is a lone pair adjacent to a C+, so we draw the curved arrow associated with that pattern (showing the lone pair becoming a bond) Notice that all three resonance structures have zero net charge: 59 (f) Recall that a lone pair next to a bond will be delocalized by resonance The lone pairs on the oxygen of the C=O bond are localized One of the lone pairs on the other oxygen atom (attached to the aromatic ring) is delocalized via resonance The lone pair on the nitrogen atom is delocalized via resonance (g) All sp2-hybridized carbon atoms are trigonal planar All sp3 hybridized carbon atoms are tetrahedral The nitrogen atom is trigonal planar The oxygen atom of the C=O bond does not have a geometry because it is connected to only one other atom, and the other oxygen atom has bent geometry (see SkillBuilder 1.8) 2.74 (a) The molecular formula is C16H21NO2 (b) Each of the highlighted carbon atoms (below) has four sigma bonds (the bonds to hydrogen are not shown) As such, these nine carbon atoms are sp3 hybridized 2.75 (a) In Section 1.5, we discussed inductive effects and we learned how to identify polar covalent bonds In this case, there are two carbon atoms that participate in polar covalent bonds (the C‒Br bond and the C‒O bond) Each of these carbon atoms will be poor in electron density (+) because oxygen and bromine are each more electronegative than carbon: (b) There are two carbon atoms that are adjacent to one or both oxygen atoms These carbon atoms will be poor in electron density (+), because oxygen is more electronegative than carbon: (c) There are seven carbon atoms that are each using a p orbital to form a bond As such, these seven carbon atoms (highlighted) are sp2 hybridized The carbon atom of the carbonyl (C=O) group is especially electron deficient, as a result of resonance (c) There are two carbon atoms that are adjacent to electronegative atoms These carbon atoms will be poor in electron density (+), because oxygen and chlorine are each more electronegative than carbon: (d) There are no sp-hybridized carbon atoms in this structure (e) There are five lone pairs (the nitrogen atom has one lone pair and each oxygen atom has two lone pairs): The carbon atom of the carbonyl (C=O) group is especially electron deficient, as a result of resonance 60 CHAPTER 2.76 We begin by drawing all significant resonance structures, and then considering the placement of the formal charges on carbon atoms in each of those resonance structures (highlighted below) A position that bears a positive charge is expected to be electron deficient (+), while a position that bears a negative charge is expected to be electron rich () The following is a summary of the electron-deficient positions and the electron-poor positions, as indicated by the resonance structures above 2.77 (a) Compound B has one additional resonance structure that compound A lacks, because of the relative positions of the two groups on the aromatic ring Specifically, compound B has a resonance structure in which one oxygen atom has a negative charge and the other oxygen atom has a positive charge: 2.78 (a) The following group is introduced, and it contains five carbon atoms: Compound A does not have a significant resonance structure in which one oxygen atom has a negative charge and the other oxygen atom has a positive charge That is, compound A has fewer resonance structures than compound B Accordingly, compound B has greater resonance stabilization (b) Compound C is expected to have resonance stabilization similar to that of compound B, because compound C also has a resonance structure in which one oxygen atom has a negative charge and the other oxygen atom has a positive charge: (b) The following highlighted carbon atom is involved in the reaction: CHAPTER (c) Compound has three significant resonance structures, shown here: The structure on the left is the most significant, because every atom has an octet and it has no formal charges The resonance hybrid is a weighted average of these three resonance structures Since the partial positive charge is delocalized onto two carbon atoms and the partial negative charge is localized on only one oxygen atom, the 61 partial negative charge is drawn larger than each of the individual partial positive charges (d) The reactive site (highlighted above) has partial positive character, which means that it is electron deficient This is what makes it reactive In the actual synthesis, this compound is treated with a carbanion (a structure containing a carbon atom with a negative charge) The reaction causes formation of a bond between the electron-deficient carbon atom and the electron-rich carbon atom We will learn about such reactions in Chapter 21 2.79 We will need to draw two resonance hybrids, one for each of the highlighted carbon atoms One highlighted position is part of an aromatic ring, and we will begin by focusing on that position In doing so, we can save time by redrawing only the relevant portion of the molecule, like this: This aromatic ring has eight significant resonance structures, shown here: 62 CHAPTER The resonance hybrid is a weighted average of these eight resonance structures Resonance structures and are equally most significant because all atoms have an octet AND there are no formal charges Resonance structures 3-8 are less significant than and but approximately equally significant to each other because every atom has a full octet with a positive charge on an oxygen atom and a negative charge on a carbon atom Since the partial negative charge is delocalized onto three carbon atoms and the partial positive charge is delocalized onto only two oxygen atoms, the partial positive charges are drawn slightly larger than the partial negative charges The analysis allows us to draw the resonance hybrid for this portion of the molecule, and it demonstrates that the highlighted carbon atom is electron rich (-) Now let’s focus on our attention on the other highlighted carbon atom (the one that is part of a carboxylic acid group) Just as we did before, we will draw only the portion of the molecule that is of interest: This portion of the molecule has four significant resonance structures, shown below: The resonance hybrid is a weighted average of these four resonance structures Resonance structure is most significant because all atoms have an octet AND there are no formal charges Resonance structure is the next most significant because there are formal charges, yet every atom has a full octet Since the partial negative charge is localized on one oxygen atom, this partial charge is the largest The partial positive charge is delocalized over three atoms, but it is larger on the oxygen (relative to either of the carbon atoms) The resonance hybrid demonstrates that this carbon atom is electron poor (+) CHAPTER 63 2.80 (a) The molecular formula for CL-20 is C6H6N12O12 The molecular formula for HMX is C4H8N8O8 (b) The lone pair is next to a bond, and therefore delocalized (see resonance structures below) 2.81 This intermediate is highly stabilized by resonance The positive charge is spread over one carbon atom and three oxygen atoms 2.82 (a) Both molecules have identical functional groups (alcohol + alkene), and both compounds have the same molecular formula (C27H46O): The structure on the left exhibits two six-membered rings and two five-membered rings, while the structure on the right has three six-membered rings and only one five-membered ring The long alkane group is apparently located in the wrong position on the five-membered ring of the incorrect structure (b) Both structures contain an alkene group, an aromatic ring, an amide group, and two ether functional groups But the incorrect structure has a third ether functional group (in the eight-membered ring), while the correct structure has an alcohol functional group The incorrect structure has an eight-membered ring, while the correct structure has a fivemembered ring The two carbon atoms and oxygen atom in the ring of the incorrect structure are not part of the ring for the correct structure 64 CHAPTER 2.83 (a) The positive charge in basic green is resonance-stabilized (delocalized) over twelve positions (two nitrogen atoms and ten carbon atoms), as seen in the following resonance structures (b) The positive charge in basic violet is expected to be more stabilized than the positive charge in basic green 4, because the former is delocalized over thirteen positions, rather than twelve Specifically, basic violet has an additional resonance structure that basic green lacks, shown here: CHAPTER In basic violet 4, the positive charge is spread over three nitrogen atoms and ten carbon atoms 2.84 Polymer contains only ester groups, so the IR spectrum of polymer is expected to exhibit a signal near 1740 cm-1 (typical for esters), associated with vibrational excitation (stretching) of the C=O bond Polymer lacks any ester groups, so the signal near 1740 cm-1 is expected to be absent in the IR spectrum of polymer Instead, polymer has OH groups, which are expected to produce a broad signal in the range 3200-3600 cm-1 Polymer has both functional groups (alcohol group and ester group), so an IR spectrum of polymer is expected to exhibit both characteristic signals When polymer is converted to polymer 4, the signal near 1740 cm-1 is expected to vanish, which would indicate complete hydrolysis of polymer In practice, the signal for the C=O stretch in polymer appears at 1733 cm-1, which is very close to our estimated value of 1740 cm-1 2.85 Compound has an OH group, which is absent in compound Therefore, the IR spectrum of should 65 exhibit a broad signal in the range 3200-3600 cm-1 (associated with O-H stretching), while the IR spectrum of would be expected to lack such a signal The conversion of to could therefore be confirmed with the disappearance of the signal corresponding with excitation of the O-H bond Another way to monitor the conversion of to is to focus on the C-H bond of the aldehyde group in compound 1, which is expected to produce a signal in the range 27502850 cm-1 Since the aldehyde group is not present in compound 2, we expect this signal to vanish when is converted to There is yet another way to monitor this reaction with IR spectroscopy Compound possesses only one C=O bond, while compound has two C=O bonds As such, the latter should exhibit two C=O signals One signal is expected to be near 1680 cm-1 (for the conjugated ketone), and the other signal should be near 1700 cm-1 (corresponding to the conjugated ester) In contrast compound has only one C=O bond, which is expected to produce a signal near 1680 cm-1 (for the conjugated aldehyde) Therefore, the conversion of to can be monitored by the appearance of a signal near 1700 cm-1 2.86 (a) This compound contains the following functional groups: (b) The nitrogen atom has a lone pair that is delocalized via resonance: In the second resonance structure shown, the C-N bond is drawn as a double bond, indicating partial double-bond character This bond is thus a hybrid between a single and double bond; the partial double bond character results in 66 CHAPTER partially restricted rotation around this bond In contrast, the C-N bond on the left experiences free rotation because that bond has only single-bond character 2.87 (a) Each of the four amides can be represented as a resonance hybrid (one example shown below) The charge-separated resonance structure indicates that there is a δ+ on the amide nitrogen, which thus pulls the electrons in the N-H bond closer to the nitrogen atom, leaving the hydrogen atom with a greater δ+ This resonance effect is not present in the NH bond of the amines Thus, the δ+ on an amide H is greater than that on an amine H, leading to a stronger hydrogen bond (b) The following intermolecular hydrogen bonds are formed during self-assembly: CHAPTER 67 2.88 (a) Anion is highly stabilized by resonance (the negative charge is delocalized over two oxygen atoms and three carbon atoms) The resonance structures for are as follows: Cation is highly stabilized by resonance (the positive charge is delocalized over two oxygen atoms and four carbon atoms) The resonance structures of are as follows: (b) Double bonds are shorter in length than single bonds (see Table 1.2) As such, the C-C bonds in compound will alternate in length (double, single, double, etc.): The double bonds have some single-bond character as a result of resonance, as can be seen in resonance structure 1c: 68 CHAPTER Similarly, the single bonds have some double-bond character, also because of resonance However, this effect is relatively small, because there is only one resonance structure (1a above) in which all atoms have an octet AND there are no formal charges Therefore, it is the most significant contributor to the overall resonance hybrid As such, the double bonds have only a small amount of single-bond character, and the single bonds have only a small amount of double-bond character In contrast, anion does not have a resonance structure that lacks charges All resonance structures of bear a negative charge Among the resonance structures, two of them (2a and 2e) contribute the most character to the overall resonance hybrid, because the negative charge is on an electronegative oxygen atom (rather than carbon) In fact, these two resonance contributors will contribute equally to the overall resonance hybrid As such, the bonds of the ring will be very similar in length, because they have both single-bond character and double-bond character in equal amounts A similar argument can be made for compound (c) In compound 1, a hydrogen bonding interaction occurs between the proton of the OH group and the oxygen atom of the C=O bond: This interaction is the result of the attraction between partial charges (+ and -) However, in cation 3, a similar type of interaction is less effective because the O of the C=O bond is now poor in electron density, and therefore less capable of forming a hydrogen bonding interaction, as can be seen in resonance structure 3a The other oxygen atom is also ineffective at forming an intramolecular hydrogen bond because it too is poor in electron density, as can be seen in resonance structure 3f: 2.89 In order for all four rings to participate in resonance stabilization of the positive charge, the p orbitals in the four rings must all lie in the same plane (to achieve effective overlap) In the following drawing, the four rings are labeled A-D Notice that the D ring bears a large substituent (highlighted) which is trying to occupy the same space as a portion of the C ring: This type of interaction, called a steric interaction, forces the D ring to twist out of plane with respect to the other three rings, like this: In this way, the overlap between the p orbitals of the D ring and the p orbitals of the other three rings is expected to be less effective As such, participation of the D ring in resonance stabilization is expected to be diminished with respect to the participation of the other three rings ... Student Study Guide and Solutions Manual, 4e for Organic Chemistry, 4e David Klein Johns Hopkins University SVP, PUBLISHING STRATEGY AND QUALITY: Elizabeth Widdicombe... organic chemistry, and use this book to check your solutions This book has also been designed to serve as a study guide, as described below WHAT’S IN THIS BOOK This book contains more than just solutions... integrated problems, and challenge problems The sections described above have been designed to serve as useful tools as you study and learn organic chemistry Good luck! David Klein Johns Hopkins